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Chapter 12: Problem 39

You need to extend a 2.50-inch-diameter pipe, but you have only a 1.00-inch- diameter pipe on hand. You make a fitting to connect these pipes end to end. If the water is flowing at 6.00 cm s in the wide pipe, how fast will it be flowing through the narrow one?

Short Answer

Expert verified
The water flows at about 37.49 cm/s in the 1.00-inch pipe.

Step by step solution

01

Identify the Known Quantities

Start by identifying the given data in the problem. You have a wide pipe with a diameter of 2.50 inches and a narrow pipe with a diameter of 1.00 inch. The flow speed in the wide pipe is 6.00 cm/s. We need to find the flow speed through the narrow pipe using this information.
02

Convert Diameters to Radii and then to Units

The diameter of the wide pipe is 2.50 inches, and the narrow pipe is 1.00 inch. First, convert these diameters into radii: \[ r_1 = \frac{2.50}{2} = 1.25 \text{ inches} \]\[ r_2 = \frac{1.00}{2} = 0.50 \text{ inches} \]Convert these radii from inches to centimeters (1 inch = 2.54 cm):\[ r_1 = 1.25 \times 2.54 = 3.175 \text{ cm} \]\[ r_2 = 0.50 \times 2.54 = 1.27 \text{ cm} \]
03

Calculate the Cross-Sectional Areas

Calculate the cross-sectional area for each pipe using the formula for the area of a circle, \(A = \pi r^2\):\[ A_1 = \pi \times (3.175)^2 = 31.664 \text{ cm}^2 \]\[ A_2 = \pi \times (1.27)^2 = 5.067 \text{ cm}^2 \]
04

Apply the Principle of Continuity

According to the principle of continuity for incompressible fluid flow, the flow rate must be constant. This means:\[ A_1 \cdot v_1 = A_2 \cdot v_2 \]Substitute the known values to find \(v_2\):\[ 31.664 \times 6.00 = 5.067 \times v_2 \]
05

Solve for the Speed in the Narrow Pipe

Rearrange the equation to solve for \(v_2\):\[ v_2 = \frac{31.664 \times 6.00}{5.067} \]Calculate:\[ v_2 = \frac{189.984}{5.067} \approx 37.49 \text{ cm/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fluid Dynamics
Fluid dynamics is a branch of physics studying how fluids (liquids and gases) move. In our scenario, we are dealing with water flowing through pipes of different diameters.

Understanding fluid dynamics involves analyzing variables such as velocity, pressure, and flow rate. The flow behavior can be described using various principles and equations, such as Bernoulli鈥檚 equation and the continuity equation.

The continuity equation tells us that for an incompressible fluid like water, the product of the cross-sectional area and the velocity at any two points in the flow should remain constant. Thus, if the fluid's cross-sectional area changes, its velocity must adjust to maintain a constant flow rate.
Cross-Sectional Area
The cross-sectional area of a pipe is key to understanding how fluid flows through it. In essence, the cross-section is the face of the pipe that fluid flows through.

It is calculated using the formula for the area of a circle, which is \(A = \pi r^2\). Here, \(r\) is the radius of the pipe. When the radius increases, the cross-sectional area grows quadratically, greatly affecting flow characteristics.
  • If the cross-sectional area is large, the fluid can flow more slowly because there is more space.
  • If the area is smaller, the fluid must flow more quickly to move the same volume.
In our problem, calculating the cross-sectional area for both the wide and narrow pipes was essential for determining how fast the water moves through each.
Flow Rate
The flow rate is the volume of fluid that travels through a cross-section of a pipe per unit time. It remains constant in an incompressible fluid flow, as the fluid cannot be compressed to occupy less space.

In any continuous flow, this flow rate can be expressed as the product of the cross-sectional area and the velocity of the liquid. Thus, we get the equation: \(Q = A \times v\) where \(Q\) is the flow rate, \(A\) is the cross-sectional area, and \(v\) is the velocity.
  • As the cross-sectional area of a pipe changes, the velocity also changes to keep the flow rate consistent.
For the given problem, even though the pipe changed in diameter, the flow rate remained identical in both sections of the pipe, which enabled us to calculate the velocity in the narrow pipe.
Pipe Diameters
Pipe diameters are a crucial factor in determining how fluid flows through a pipe system. They directly affect the cross-sectional area. The larger the diameter, the larger the cross-sectional area will be.

Pipe diameter changes impact the velocity of fluid inside:
  • Larger diameters mean a slower fluid velocity if the flow rate is constant, due to a larger cross-section allowing more fluid through at once.
  • Conversely, a smaller diameter means a faster velocity for the same reason鈥攊t restricts space, requiring the fluid to speed up to maintain flow rate consistency.
In our scenario, the diameter of the wide pipe was 2.50 inches, while the narrow one was only 1.00 inch. This significant difference required the water to speed up considerably as it transitioned from the wide to the narrow pipe, as calculated using the principle of continuity.

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Most popular questions from this chapter

The weight of a king's solid crown is w. When the crown is suspended by a light rope and completely immersed in water, the tension in the rope (the crown's apparent weight) is (a) Prove that the crown's relative density (specific gravity) is 1\(/(1-f)\) . Discuss the meaning of the limits as \(f\) approaches 0 and \(1 .\) (b) If the crown is solid gold and weighs 12.9 \(\mathrm{N}\) in air, what is its apparent weight when completely immersed in water? (c) Repeat part (b) if the crown is solid lead with a very thin gold plating, but still has a weight in air of 12.9 \(\mathrm{N}\) .

A hunk of aluminum is completely covered with a gold shell to form an ingot of weight 45.0 N. When you suspend the ingot from a spring balance and submerge the ingot in water, the balance reads 39.0 N. What is the weight of the gold in the shell?

A cube 5.0 \(\mathrm{cm}\) on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 \(\mathrm{cm}\) in diameter all the way through and perpendicular to one face, you find that the cube weighs 7.50 \(\mathrm{N}\) . (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

A barge is in a rectangular lock on a freshwater river. The lock is 60.0 \(\mathrm{m}\) long and 20.0 \(\mathrm{m}\) wide, and the steel doors on each end are closed. With the barge floating in the lock, a \(2.50 \times 10^{6} \mathrm{N}\) load of scrap metal is put onto the barge. The metal has density 9000 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?

The densities of air, helium, and hydrogen (at \(p=1.0\) atm and \(T=20^{\circ} \mathrm{C}\) ) are \(1.20 \mathrm{kg} / \mathrm{m}^{3}, 0.166 \mathrm{kg} / \mathrm{m}^{3},\) and \(0.0899 \mathrm{kg} / \mathrm{m}^{3},\) respectively. (a) What is the volume in cubic meters displaced by a hydrogen-filled airship that has a total "lift" of 90.0 \(\mathrm{kN}\) ? (The "lift" is the amount by which the buoyant force exceeds the weight of the gas that fills the airship.) (b) What would be the "lift" if helium were used instead of hydrogen? In view of your answer, why is helium used in modern airships like advertising blimps?
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